Chapter 12, Problem 30A

Summary Introduction

To identify:

The mRNA sequence and orientation from the given sequence in non- template strand.

Introduction:

The central dogma of biology is that the DNA codes for RNA which guides the synthesis of proteins. RNA is nucleic acid similar to DNA but in RNA the base uracil replaces thymine, and it is single stranded. Three major types of RNA re found in living cells; messenger RNA (mRNA), ribosomal RNA (rRNA) and transfer RNA (tRNA).

Answer to Problem 30A

The mRNA sequence would be 5’AUG CCA GUC AUC 3’.

The amino acid sequence is methionine- proline- valine- isoleucine.

Explanation of Solution

Each DNA molecule consists of two strands. Both the strands have complementary base pairs. During transcription, the two strands of DNA unwind. One strand serves as a template to synthesize new mRNA molecule while the other strand that remains unused is called the non- template strand.

It is given that the sequence in non- template strand is as follows:

5’ATGCCAGTCATC 3’ The sequence in template strand would be complementary to the sequence in non- template strand. Its orientation will be from 3’ to 5’.

Hence, sequence in template strand would be 3’TACGGTCAGTAG 5’.

Through transcription, the DNA code is transferred to mRNA. The mRNA transcript is manufactured in a 5’ to 3’ direction adding each new nucleotide to the 3’ end. Free RNA nucleotides base pairs with their complementary nucleotides on the DNA strand. Uracil is incorporated instead of thymine as the mRNA molecule is made.

So the mRNA sequence would be 5’AUGCCAGUCAUC 3’.

The dictionary of genetic code helps in finding the amino acids coded by the codons in mRNA.

AUG codes for methionine

CCA codes for proline

GUC codes for valine

AUC codes for isoleucine

Hence, the amino acid sequence is methionine- proline- valine- isoleucine